Why do function composition and application have a dependent implementation in Agda?

Question

Why do function composition (∘) and application ($) have the implementation as available in https://github.com/agda/agda-stdlib/blob/master/src/Function.agda#L74-L76? Copied here for convenience:

_∘_ : ∀ {a b c}
        {A : Set a} {B : A → Set b} {C : {x : A} → B x → Set c} →
        (∀ {x} (y : B x) → C y) → (g : (x : A) → B x) →
        ((x : A) → C (g x))
f ∘ g = λ x → f (g x)

_∘'_ : ∀ {a b c} {A : Set a} {B : Set b} {C : Set c} →
         (B → C) → (A → B) → (A → C)
f ∘' g = λ x → f (g x)

_$_ : ∀ {a b} {A : Set a} {B : A → Set b} →
      ((x : A) → B x) → ((x : A) → B x)
f $ x = f x

_$'_ : ∀ {a b} {A : Set a} {B : Set b} →
       (A → B) → (A → B)
f $' x = f x

I initially thought the rationale behind this was that $ would be able to handle higher order types that $' wouldn't be able to handle. For example, consider A=Nat, B=List, f is ::, where B depends on A. But after a lot of testing, I couldn't come up with an example that would show that the implementation of $' is not sufficient. What scenarios does $ handle that $' isn't able to handle? (Similarly, what scenarios does ∘ handle that ∘' doesn't?

open import Agda.Builtin.Nat public
open import Agda.Primitive public

--data List {a} (A : Set a) : Set a where
--  []  : List A
--  _∷_ : (x : A) (xs : List A) → List A

data Vec {a} (A : Set a) : Nat → Set a where
  []  : Vec A zero
  _∷_ : ∀ {n} (x : A) (xs : Vec A n) → Vec A (suc n)

tail : ∀ {a n} {A : Set a} → Vec A (suc n) → Vec A n
tail (x ∷ s) = s

_$_ : ∀ {a b} {A : Set a} {B : A → Set b} →
      ((x : A) → B x) → ((x : A) → B x)
f $ x = f x

_$'_ : ∀ {a b} {A : Set a} {B : Set b} →
       (A → B) → (A → B)
f $' x = f x

_∘_ : ∀ {a b c}
        {A : Set a} {B : A → Set b} {C : {x : A} → B x → Set c} →
        (∀ {x} (y : B x) → C y) → (g : (x : A) → B x) →
        ((x : A) → C (g x))
f ∘ g = λ x → f (g x)

_∘'_ : ∀ {a b c} {A : Set a} {B : Set b} {C : Set c} →
         (B → C) → (A → B) → (A → C)
f ∘' g = λ x → f (g x)

Vecc : ∀ {a} → Nat → (A : Set a) → (Set a)
Vecc x y = Vec y x

data Pair {a b} (A : Set a) (B : A → Set b) : Set (a ⊔ b) where
  _,_ : (x : A) → (y : B x) → Pair A B

-- Dependent Pair attempt
--fst : ∀ {a b} {A : Set a} {B : A → Set b} → Pair A B → A
--fst (a , b) = a
--
--f : Pair Nat $' Vec Nat
--f = _,_ zero $' []
--
--g : Pair (Pair Nat $' Vec Nat) $' λ x → Nat
--g = _,_ (_,_ zero $' []) $' zero

-- Some other attempt
--f : ∀ {a n} {A : Set a} → Vec A ((suc ∘' suc) n) → Vec A n
--f {a} = tail {a} ∘' tail {a}

-- Vec attempt
--f : ∀ {a} (A : Set a) → (Set a)
--f {a} = Vecc {a} (suc zero) ∘' Vecc {a} (suc zero)
--
--h = f Nat
--
--x : h
--x = (zero ∷ []) ∷ []

-- List attempt
--f : ∀ {a} (A : Set a) → (Set a)
--f {a} = List {a} ∘' List {a}
--
--g : ∀ {a} (A : Set a) → (Set a)
--g {a} = List {a} ∘ List {a}
--
--h = f Nat
--i = g Nat
--
--x : h
--x = (zero ∷ []) ∷ []

Answer 1

One works with dependent functions, the other doesn't, as Andras Kovacs mentioned.

The important difference is that for non-dependent functions stronger proofs can be constructed. For example:

eq : {A B} -> f : (A -> B) -> x y : A -> x == y -> (f x) == (f y)
eq f x .x refl = refl

Here we can construct equality of f x and f y. But we can't do the same for dependent functions - because there is no way to prove B x == B y. So there is only a weaker proof that f x can be "cast" to f y.

transport : {A} {B : A -> Set} -> f : (x : A -> B x) -> x y : A -> x == y -> f x -> f y
transport f x .x refl fx = fx

(Actually, transport is usually defined as B x -> B y, not for a dependent function; but I just can't come up with a better name)

Answer 2

∘′ and $′ don't work with dependent functions. You simply didn't try any tests with dependent functions. For f $ x examples, f must be dependent, for f ∘ g, either of the functions must be dependent. Example:

open import Data.Nat
open import Data.Vec
open import Function
open import Relation.Binary.PropositionalEquality

replicate' : {A : Set} → A → (n : ℕ) → Vec A n
replicate' a n = replicate a

refl' : {A : Set}(a : A) → a ≡ a
refl' a = refl

-- fail1 : Vec ℕ 10
-- fail1 = replicate' 10 $′ 10

ok1 : Vec ℕ 10
ok1 = replicate' 10 $ 10

-- fail2 : ∀ n → replicate' 10 n ≡ replicate' 10 n
-- fail2 = refl' ∘′ replicate' 10

ok2 : ∀ n → replicate' 10 n ≡ replicate' 10 n
ok2 = refl' ∘ replicate' 10